How do I find all points on a curve whose x coordinate is 1? Then, how do I write an equation for the tangent line at each of these points?

The curve is xy^2-x^3y=6

& I already have the derivative.

To answer the first question:

Plug in $\displaystyle x=1$ into the original equation to find the y values.

$\displaystyle y^2-y=6\implies y^2-y-6=0$

This can be easily factored. You will get two y values, which I will denote as $\displaystyle y_1$ and $\displaystyle y_2$ respectively.

This will give you the two points: $\displaystyle (1,y_1)$ and $\displaystyle (1,y_2)$

Now, find the derivative [which you already did] and then plug these points into $\displaystyle \frac{\,dy}{\,dx}$. Each point will create a different slope, $\displaystyle m_1$ and $\displaystyle m_2$, respectively.

Now to find the equations of the tangents, use the point slope form of a line. You will have two equations:

Plug in $\displaystyle x=1$ into the original equation to find the y values.

$\displaystyle y^2-y=6\implies y^2-y-6=0$

This can be easily factored. You will get two y values, which I will denote as $\displaystyle y_1$ and $\displaystyle y_2$ respectively.

This will give you the two points: $\displaystyle (1,y_1)$ and $\displaystyle (1,y_2)$

Now, find the derivative [which you already did] and then plug these points into $\displaystyle \frac{\,dy}{\,dx}$. Each point will create a different slope, $\displaystyle m_1$ and $\displaystyle m_2$, respectively.

Now to find the equations of the tangents, use the point slope form of a line. You will have two equations:

$\displaystyle y-y_1=m_1(x-1)$ and $\displaystyle y-y_2=m_2(x-1)$

Can you take it from here?

--Chris

You are AMAZING.
One last thing, how would I find the x-coordinate of each point on the cure where the tangent line is vertical??